With (2) you start off "since is divisible by 4" but that is what you are trying to prove. You have to start by proving it is true for n= 1 (which is easy) and then say "If, for some k, is divisible by 4" or "suppose is divisible by 4". The difference is that (1) I am saying "if" rather than "since" and (2) I am using "k" rather than "n". That is not crucially important but makes it clear that I am not assuming what I want to prove.

For (3), I personally don't like using "formulas" like that. Here is what I would do:
Assume a population of 10000 people. 4%, 400, use drugs and 96%, 9600, do not. There is a test for drug use which has a 3% false positive rate and a 2% false negative rate. So of the 9600 who do not use drugs, 97%, 9312, will test negative and 3%, 288, will test positive. Of the 400 who use drugs, 98%, 392, test positive and 2%, 8, test negative.

So there are a total of 288+ 392= 680 who test positive, 392 of whom are really drug users. So the probability that a person who tests positive really is a drug user is [tex]\frac{392}{680}= .5765... or 57.65% as you have.

Similarly, there are a total of 9312+ 8= 9320 who test negative, 9312 of whom do not use drugs. So the probability that a person who tests negative is not a drug user is or 99.9% as you have.

Not part of the proof: This means that there certainly are one or more positive integers such that if n is one of those integers 5^n - 1 is evenly divisible by 4. Choose an arbitrary one of those integers and call it k. This then, at least to me, completely justifies the induction hypothesis as being more than an arbitrary assumption. Back to the proof.

Okay for step 1 did you mean n=1 => 5^1 -1 = 5 - 1 = 4 = 4*1 => 5^n -1 instead of just 5n is evenly divisible by 4?
Does it matter if we use that base case or would the base case of n = 0, 5^0 -1 = 1 -1 = 0 and 4 is divisible by 4 work better?

Okay for step 1 did you mean n=1 => 5^1 -1 = 5 - 1 = 4 = 4*1 => 5^n -1 instead of just 5n is evenly divisible by 4?
Does it matter if we use that base case or would the base case of n = 0, 5^0 -1 = 1 -1 = 0 and 4 is divisible by 4 work better?

I personally think saying that is an assumption is silly because in step 1 you have already proved that at least one such positive integer exists, but make sure your teacher agrees.

B. I think the word "arbitrary" is important in my style. You know only that k has the properties common to all positive integers and the property that you are trying to prove is general. So k could be any of the positive integers for which the property in question is true. In your proof, you said "k is a positive integer such that ..." I prefer "k is an arbitrary positive integer such that."

C. I am pretty sure a fussy grader would object to your "and m is a positive integer ..." because, for those who think that you are making an assumption about k, you SEEM to be making an extra assumption about m. In fact, the existence of m is a consequence of what you have already affirmed about k. If 5^k - 1 is evenly divisible by 4, then, by the meaning of the phrase "evenly divisible by 4," there exists a positive integer m such that 4m = 5^k - 1. So I would say

Again, this way of showing a proof by weak mathematical induction strikes me personally as being intuitive, but your teacher may not like it because it is a little atypical. So check with your teacher about what he or she likes.